Question

In the following compounds, what is the increasing order of their reactivity towards nucleophilic addition reactions?

• A. Benzaldehyde $<$ p-Tolualdehyde $<$ p-Nitrobenzaldehyde $<$ Acetophenone
• B. Acetophenone $<$ Benzaldehyde $<$ p-Tolualdehyde $<$ p-Nitrobenzaldehyde
• C. Benzaldehyde $<$ Acetophenone $<$ p-Tolualdehyde $<$ p-Nitrobenzaldehyde
• D. Acetophenone $<$ p-Tolualdehyde $<$ Benzaldehyde $<$ p-Nitrobenzaldehyde

Answer 1

The Correct Option is D

As discussed in the previous response, the reactivity towards nucleophilic addition reactions depends on the electronic and steric effects of the substituents on the carbonyl group.

Let's reiterate the analysis:

Benzaldehyde (C₆H₅CHO): A simple aromatic aldehyde.

p-Tolualdehyde (p-CH₃C₆H₄CHO): The methyl group (CH₃) is electron-donating, decreasing reactivity.

p-Nitrobenzaldehyde (p-NO₂C₆H₄CHO): The nitro group (NO₂) is electron-withdrawing, increasing reactivity.

Acetophenone (C₆H₅COCH₃): A ketone with steric hindrance and electron donation from the methyl group, decreasing reactivity.

The electron-withdrawing nitro group in p-nitrobenzaldehyde makes it the most reactive, while the electron-donating methyl group in p-tolualdehyde reduces reactivity. The ketone, acetophenone, is the least reactive due to steric hindrance and electron donation from the methyl group. Benzaldehyde is intermediate in reactivity.

Therefore, the increasing order of reactivity is:

Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde

The correct answer is:

Option 4: Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde